上海青浦区2022届九年级初三数学一模试卷+答案

2022年上海市青浦区中考数学一模试卷2022.1一、选择 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 ：（本大题共6题，每小题4分，满分24分）[每题只有一个正确选项，在答题纸相应题号的选项上用2B铅笔正确填涂]1.下列图形，一定相似的是（）（A）两个直角三角形；（B）两个等腰三角形；（C）两个等边三角形；（D）两个菱形．ll2.如图，已知AB//CD//EF，它们依次交直线1、2于点A、C、E和点B、D、F．如果AC∶CE=2∶3，BD=4，那么BF等于（）（A）6；（B）8；（C）10；（D）12．（第2题图）3.在Rt△ABC中，∠C=90º，那么cotA等于（）ACACBCBC（A）；（B）；（C）；（D）．BCABACAB4.如图，点D、E分别在△ABC的边AB、BC上，下列条件中一定能判定DE∥AC的是（）（第4题图）ADBEBDBEADCEBDDE（A）；（B）；（C）；（D）．DBCEADECABBEBAAC5.如果a2b（a、b均为非零向量），那么下列结论错．误．的是（）（A）|a|2|b|；（B）a∥b；（C）a2b0；（D）a与b方向相同．6.如图，在平行四边形ABCD中，点E在边BA的延长线上，联结EC，交边AD于点F，则下列结论一定正确的是（）EAAFEAFDAFEAEAAF（A）；（B）；（C）；（D）．ABBCABAFBCCDEBAD（第6题图）二、填空题：（本大题共12题，每小题4分，满分48分）[请将结果直接填入答题纸的相应位置]7.已知线段b是线段a、c的比例中项，且a=1，b=3，那么c=▲．8.计算：3a2(a2b)=▲．9.如果两个相似三角形的周长比为2∶3，那么它们的对应高的比为▲．10.二次函数yx2x1的图像有最▲点．（填“高”或“低”）11.将抛物线yx2向下平移2个单位，所得抛物线的 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 达式是▲．12.如果抛物线yax2bxc（其中a、b、c是常数，且a≠0）在对称轴左侧的部分是下降的，那么a▲0．（填“＜”或“＞”）13.在△ABC中，∠C=90º，如果tan∠A=2，AC=3，那么BC=▲．14.如图，点G为等边三角形ABC的重心，联结GA，如果AG=2，那么BC=▲．（第15题图）（第16题图）（第14题图）15.如图，如果小华沿坡度为1:3的坡面由A到B行走了8米，那么他实际上升的高度为▲米．16.如图，在边长相同的小正方形组成的网格中，点A、B、O都在这些小正方形的顶点上，那么sin∠AOB的值为▲．17.如图，在矩形ABCD中，∠BCD的角平分线CE与边AD交于点E，∠AEC的角平分线与边CB的延长线交于点G，与边AB交于点F，如果AB=32，AF=2BF，那么GB=▲．18.如图，一次函数yaxb(a0，b0)的图像与x轴，y轴分别相交于点A，点B，将它绕点O逆时针旋转90°后，与x轴相交于点C，我们将图像过点A，B，C的二次函数叫做与这个一次函数关联的二次函数．如果一次函数ykxk(k0)的关联二次函数是ymx22mxc（m0），那么这个一次函数的解析式为▲．（第17题图）（第18题图）三、解答题（本大题共7题，满分78分）[请将解题过程填入答题纸的相应位置]19.（本题满分10分）01计算：sin451+2cos30tan60cot60．20.（本题满分10分,第（1）小题5分，第（2）小题5分）如图，在平行四边形ABCD中，点E在边AD上，CE、BD相交于点F，BF=3DF．（1）求AE∶ED的值；（2）如果DCa，EAb，试用a、b表示向量CF．21.（本题满分10分,第（1）小题5分，第（2）小题5分）1如图，在△ABC中，点D是BC的中点，联结AD，AB=AD，BD=4，tanC．4（1）求AB的长；（2）求点C到直线AB的距离．（第21题图）22.（本题满分10分）如图，某校的实验楼对面是一幢教学楼，小张在实验楼的窗口C（AC∥BD）处测得教学楼顶部D的仰角为27°，教学楼底部B的俯角为13°，量得实验楼与教学楼之间的距离AB=20米．求教学楼BD（BD⊥AB）的高度．（精确到0.1米）（参考数据：sin13°≈0.22，cos13°≈0.97，tan13°≈0.23，sin27°≈0.45，cos27°≈0.89，tan27°≈0.51）（第22题图）23.（本题满分12分，第（1）小题6分，第（2）小题6分）已知：如图，在四边形ABCD中，AC、BD相交于点E，∠ABD=∠CBD，DC2DEDB．（1）求证：△AEB∽△DEC；（2）求证：BCADCEBD．（第23题图）24.（本题满分12分，其中第（1）小题4分，第（2）小题4分，第（3）小题4分）如图，在平面直角坐标系xOy中，抛物线yx2bxc与x轴交于点A（-1，0）和点B（3，0），与y轴交于点C，顶点为点D．（1）求该抛物线的表达式及点C的坐标；（2）联结BC、BD，求∠CBD的正切值；（3）若点P为x轴上一点，当△BDP与△ABC相似时，求点P的坐标．（第24题图）（备用图）25.（本题满分14分，其中第（1）小题4分，第（2）小题4分，第（3）小题6分）在四边形ABCD中，AD∥BC，AB=5，AD=2，DC=25，tan∠ABC=2（如图）．点E是射线AD上一点，点F是边BC上一点，联结BE、EF，且∠BEF=∠DCB．（1）求线段BC的长；（2）当FB=FE时，求线段BF的长；（3）当点E在线段AD的延长线上时，设DE=x，BF=y，求y关于x的函数解析式，并写出x的取值范围．（第25题图）（备用图）2022年上海市青浦区中考数学一模试卷 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 一、选择题：1．C；2．C；3．A；4．B；5．D；6．D．二、填空题：27．9；8．a4b；9．2:3；10．高；11．yx2；12．；1013．6；14．23；15．4；16．10；17．22；18．y3x+3．三、解答题：123031+2322319．解：原式=．···········································（4分）21+313=2．······························································（4分）2=2．·················································································（2分）20．解：（1）∵四边形ABCD是平行四边形，∴AD//BC，AD=BC．··································································（2分）BCBF∴EDDF．·········································································（1分）BF3∵BF=3DF，∴DF．BC3∴ED．·············································································（1分）AD3∴ED．∴AE∶ED=2．··········································································（1分）1DEEA（2）∵AE∶ED=2∶1，∴2．∵EAb，1DEb∴2．···········································································（1分）∵CEDEDC，1CEba∴2．······································································（1分）CFBF∵AD//BC，∴CEBD．························································（1分）BF3CF3∵BF=3DF，∴BD4．∴CE4．3CFCE∴4．········································································（1分）3133CFbaba∴4284．·················································（1分）21．解：（1）∵过点A作AH⊥BD，垂足为点H．∵AB=AD，∴BH=HD．·····························································（1分）∵点D是BC的中点，∴BD=CD．∵BD=4，∴CD=4．∴HC=6．·················································································（1分）1AH13tanCAH∵4，∴HC4，∴2．······································（1分）22∵ABBHAH，2235AB222∴．··························································（2分）（2）过点C作CG⊥BA，交BA的延长线于点G．······························（1分）AHCGsinB∵ABBC，·····························································（2分）3CG258∴2．·········································································（1分）24CG∴5．24∴点C到直线AB的距离为5．··················································（1分）22．解：过点C作CH⊥BD，垂足为点H．·····················································（1分）由题意，得∠DCH=27°，∠HCB=13°，AB=CH=20（米）．DHtanDCH在Rt△DHC中，∵CH，∴DHtan272010.2．······（4分）HBtanHCB在Rt△HCB中，∵CH，∴BHtan13204.6．·········（4分）∴BD=HD+HB10.2+4.6=14.8（米）．·················································（1分）答：教学楼BD的高度约为14.8米．223．证明：（1）∵DCDEDB，DCDB∴DEDC．······································································（1分）又∵∠CDE=∠BDC，∴△DCE∽△DBC．·····································（1分）∴∠DCE=∠DBC．···································································（1分）∵∠ABD=∠DBC，∴∠DCE=∠ABD．···································································（1分）又∵∠AEB=∠DEC，∴△AEB∽△DEC．····································（2分）AEDE（2）∵△AEB∽△DEC，∴EBEC．·············································（1分）又∵∠AED=∠BEC，∴△AED∽△BEC．····································（1分）∴∠ADE=∠BCE．····································································（1分）又∵∠ABD=∠DBC，∴△BDA∽△BCE．····································（1分）BDDA∴BCCE．········································································（1分）∴BCADCEBD．···························································（1分）224．解：（1）将A（-1，0）、B（3，0）代入yx+bx+c，得1bc0，b2，93bc0.c3.解得：············································（2分）2所以，yx2x3．····························································（1分）当x=0时，y3．∴点C的坐标为（0，-3）．······························（1分）2yx22x3=x14（2）∵，∴点D的坐标为（1，-4）．·············（1分）∵B（3，0）、C（0，-3）、D（1，-4），∴BC=32，DC=2，BD=25．222∴BC+DC18220DB．············································（1分）∴∠BCD=90°．·········································································（1分）DC21∴tan∠CBD=BC323．·······················································（1分）AO1（3）∵tan∠ACO=OC3，∴∠ACO=∠CBD．····································（1分）∵OC=OB，∴∠OCB=∠OBC=45°．∴∠ACO+∠OCB=∠CBD+∠OBC．即：∠ACB=∠DBO．·······························································（1分）∴当△BDP与△ABC相似时，点P在点B左侧．ACDB（i）当CBBP时，1025∴32BP．∴BP=6．∴P（-3，0）．·······························（1分）ACBP（ii）当CBDB时，10BP101∴3225．∴BP=3．∴P（-3，0）．···························（1分）1综上，点P的坐标为（-3，0）或（-3，0）．25．解：（1）过点A、D分别作AH⊥BC、DG⊥BC，垂足分别为点H、点G．可得：AD=HG=2，AH=DG．∵tan∠ABC=2，AB=5，∴AH=2，BH=1．·······································································（2分）∴DG=2．22∵DC=25，∴CG=DCDG4．··········································（1分）∴BC=BH+HG+GC=1+2+4=7．······················································（1分）（2）过点E作EM⊥BC，垂足为点M．可得EM=2．DG1GC由（1）得，tan∠C=2．∵FB=FE，∴∠FEB=∠FBE．∵∠FEB=∠C，∴∠FBE=∠C．····················································（1分）1EM1BM∴tan∠FBE=2．∴2，∴BM=4．·······································（1分）2222224FB2FB∵FMEMFE，∴．······················（1分）5∴BF=2．···············································································（1分）（3）过点E作EN//DC，交BC的延长线于点N．∵DE//CN，∴四边形DCNE是平行四边形．∴DE=CN，∠DCB=∠ENB．∵∠FEB=∠DCB，∴∠FEB=∠ENB．············································（1分）又∵∠EBF=∠NBE，∴△BEF∽△BNE．···································································（1分）BFBE2∴BEBN．∴BEBFBN．··············································（1分）过点E作EQ⊥BC，垂足为点Q．可得EQ=2，BQ=x+3．2BE2QE2BQ2x322=x26x13∴．·······················（1分）y7xx26x13∴．21145x6x130xy2∴7x．······································（2分）